Run static calculations with off-nominal voltages

To run a grid on an off-nominal voltage you need to...

• adjust the operational voltage of all network feeders, that feed the grid in an off-nominal voltage level.
• adjust the turns ratio on the primary and/or secondary side of the transformers, depending which side(s) is/are operated off-nominal.

Example

In this simple example, we have a 20kV medium voltage grid, operated at 22kV, fed by a 110kV high voltage grid, operated at 104kV. The screenshot shows, how the network feeder operational voltage and the transformer turns ratios have to be calculated, such that the operational voltages are met in a power flow calculation. To play around yourself, you can download our example Grid.

Manual adjustment of Operational voltage

Setup

In this example the grid model is set up as follows:

• Nominal voltage on the high voltage side is $V_{nom}^{HV}=10kV$.
• Nominal voltage on the low voltage side is $V_{nom}^{LV}=0.4kV$.
• The network feeder operates at an off-nominal voltage of $V^{HV}=10.5kV$, which is equivalent to a relative voltage of $v^{HV}=1.05pu$.
• The rated voltages of the transformer (on the center tap position) are $V_R^{HV}=11kV$ and $V_R^{LV}=0.415kV$. The transformer voltage specifications are summarized in the following two tables:

	Rated Voltage $V_R$ [kV]
HV	11 +- 2x2.5%
LV	0.415

Tap no.	HV voltage [kV]
1	11.550
2	11.275
3	11.000
4	10.725
5	10.450

The grid with the above specifications can be downloaded and imported into Adaptricity.

Definitions

The following definitions will be required for the calculations that will follow in the next sections:

Transformer Voltage Ratio

The voltage ratio of a transformer is defined as the ratio of the primary voltage (high voltage) to the secondary voltage (low voltage):

$N=\frac{V_R^{HV}}{V_R^{LV}}$

Since in power systems we usually work with per unit (pu) values, we need to rewrite the above equation to incorporate the nominal voltages $V_{nom}^{HV}$ and $V_{nom}^{LV}$.

The voltage ratio with per unit voltage values can then be written as:

$n=\frac{V_R^{HV}}{V_R^{LV}}\cdot \frac{V_{nom}^{LV}}{V_{nom}^{HV}}=\frac{r^{LV}}{r^{HV}}$

where $r^{HV}$ we call the primary turns ratio and $r^{LV}$ we call the secondary turns ratio. These are the ratios of nominal voltage $V_{nom}$ to rated voltage $V_R$ of the transformer on the high and low voltage side respectively.

The voltage ratio can also be written as

$n\approx \frac{v^{HV}}{v^{LV}}$

where $v^{HV}$ and $v^{LV}$ are the per unit operational voltages on the high voltage and low voltage side of the transformer. Note that the equality only holds under approximation because we assume a lossless transformer for these examples.

From both equations of $n$ it follows that

$\frac{r^{LV}}{r^{HV}}\approx \frac{v^{HV}}{v^{LV}}$

We will use this equation in the following sections.

INFO

In Adaptricity, we cannot directly change the rated voltage parameters $V_R^{HV}$ and $V_R^{LV}$ of the transformer. We can only change $r^{HV}$ and $r^{LV}$.

Power Flow

Running a Power Flow on the grid specified above, we can see that under load, the voltage at the low voltage (secondary) side of the transformer drops to $v^{LV}=0.985pu$ or $V^{LV}=0.394kV$, which is $1.5\mathrm{\%}$ below the nominal voltage of $0.4kV$ (or $1pu$). This leads to voltage issues further down in the grid.

To increase the voltage on the low voltage side of the transformers, there are several options:

• adjust the operational voltage $v^{HV}$ of the network feeder
• adjust the operational voltage of the transformer by changing the rated voltage $V_R^{HV}$ on the high voltage (primary) side of the transformer.

In the following, both options will be presented in detail. The goal is to achieve a voltage of $v^{LV}=1pu$ at the low voltage side of the transformer under the same load situation as above.

Adjusting the operational voltage of the network feeder

Assuming the rated voltages $V_R^{HV}$ and $V_R^{LV}$ of the transformer are not changed (i.e. the tap number remains in the center), the only way to increase voltage $v^{LV}$ on the low voltage side is to increase the voltage $v^{HV}$ at the network feeder.

For the setup above, we can calculate $r^{LV}$ and $r^{HV}$ as

$r^{LV}=\frac{V_{nom}^{LV}}{V_R^{LV}}=\frac{0.4}{0.415}=0.964$

$r^{HV}=\frac{V_{nom}^{HV}}{V_R^{HV}}=\frac{10}{11}=0.909$

and therefore

$n=\frac{0.964}{0.909}=1.06$

From the second equation for $n$ it follows that

$n\approx \frac{v^{HV}}{v^{LV}}\stackrel{!}{=}1.06$

Solving for $v^{HV}$ we get

$v^{HV}=v^{LV}\cdot 1.06=1\cdot 1.06=1.06$

We can now update the network feeder operational voltage as part of a Grid Upgrade and re-run the Power Flow.

As can be seen, we were able to increase $v^{LV}$ from $0.985pu$ to $0.994pu$.

INFO

With the simple equations above, we will not reach the desired $1pu$ exactly, due to the non-linear behaviour of the grid and the non-ideal behaviour of the transformer.

Adjusting the transformer tap position

We now assume that the operational voltage of the network feeder $v^{HV}$ is fixed. The only parameter that we can change to influence the voltage on the low voltage side is the rated voltage $V_R^{HV}$ on the high voltage side of the transformer. From the setup we can see the tap number and the corresponding rated voltage on the high voltage side of the transformer.

To achieve a voltage increase of $1.5\mathrm{\%}$ on the low voltage side, we would need to decrease the rated voltage of the transformer on the high voltage by $1.5\mathrm{\%}$. The closest tap position is number 4, which would decrease the transformer rated voltage by $2.5\mathrm{\%}$ to $V_R^{HV}=10.725kV$.

We can now calculate the primary turns ratio $r^{HV}$ that corresponds to $V_R^{HV}$ as

$r^{HV}=\frac{V_{nom}^{HV}}{V_R^{HV}}=\frac{10}{10.725}=0.9324$

We can now update the primary turns ratio as part of a Grid Upgrade and re-run the Power Flow.

As can be seen, we were able to increase $v^{LV}$ from $0.985pu$ to $1.01pu$. There is a small overshoot in voltage, due to there not being a tap position corresponding to the $1.5\mathrm{\%}$ increase required.

We can calculate the exact rated voltage $V_R^{HV}$ required to achieve $1pu$ on the low voltage side with

$V_R^{HV}=11kV\cdot (1-1.5\mathrm{\%})=10.835kV$

This results in a primary turns ratio of $r^{HV}=0.923$.

Running another Power Flow with the updated parameter, we see that the voltage $v^{LV}$ on the low voltage side is now exactly $1pu$.